Section 3 Sequences and Limits, Continued.

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1 Section 3 Sequences and Limits, Continued. Lemma 3.6 Let {a n } n N be a convergent sequence for which a n 0 for all n N and it α 0. Then there exists N N such that for all n N. α a n 3 α In particular this result ensures that if the it is non-zero then, for all sufficiently large n, the a n are not too close to 0. Proof Take ε = α / > 0 in the definition that {a n } n N has it α. The definition tells us that there exists N N such that for all n N. Thus a n α < α a n = α (α a n α + α a n 3 α by the triangle inequality, while a n = α (α a n α α a n α by Corollary.. If {a n } n N and {b n } n N are sequences we can form a new sequence {a n + b n } n N whose n-th term is a n + b n for all n N. Similarly we can form other, new, sequences {a n b n } n N and {a n b n} n N etc. Theorem 3.7 Let {a n } n N and {b n } n N be convergent sequences with n a n = α and n b n = β. Then (i {a n + b n } n N is convergent and n (a n + b n = α + β, (ii {a n b n } n N is convergent and n (a n b n = αβ, { (iii If a n 0 for all n N and α 0 then ( n a n =. α Proof Rough work a n }n N is convergent and (i Here we will examine (a n + b n (α + β = (a n α + (b n β a n α + b n β by the triangle inequality. So if we make both terms on

2 the right hand side less than ε/ then the left hand side will be less that ε as required. (ii Here we will examine a n b n αβ = a n b n αb n + αb n αβ = (a n αb n + α(b n β b n a n α + α b n β. Again we will make both terms on the RHS less than ε/. For this we will have to take account of b n (bounded since the sequence is convergent and α. (iii Here we will examine a n α = α a n a n α. We can make the numerator, α a n, small but we have to make sure that the denominator, in particular a n, does not also get too small. This, though, will follow from Lemma 3.6. End of rough work. Let ε > 0 be given. (i By definition, n a n = α implies that there exists N N such that a n α < ε/ for all n N. Similarly, n b n = β implies that there exists N N such that b n β < ε/ for all n N. Choose N = max(n, N. Then for n N we have, from the rough work above, (a n + b n (α + β a n α + b n β < ε + ε = ε. So we have verified the conditions of the definition that allow us to conclude n (a n + b n = α + β. (ii Since {b n } n N is convergent it is bounded (Theorem 3. so there exists M > 0 such that b n M for all n. Further, by taking M larger if necessary we can assume that M > α also. It is important that M is not 0.

3 Then, since n a n = α we have that there exists N 3 N such that a n α < ε M for all n N 3. Also, n b n = β implies that there exists N 4 N such that b n β < ε M for all n N 4. Choose N = max(n 3, N 4. Then, for n N, we have, from the rough work above, Hence n (a n b n = αβ. a n b n αβ b n a n α + α b n β M a n α + α b n β < M ε M + α ε M = ε + ε, since α < M, < ε, (iii Since α 0 we can apply Lemma 3.6 to {a n } n N to find N 5 such that α < a n for all n N 5. In which case a n α = α a n a n α < α a n α holds for all such n. Now, since n a n = α we can find N 6 N such that a n α < ε α for all n N 6. Choose N = max (N 5, N 6. Then for all n N we have a n α < α a n < α α ε α = ε. Hence n ( a n = α. 3

4 Corollary 3.8 Let {a n } n N and {b n } n N be convergent sequences with n a n = α and n b n = β. Then (i For any λ R, {λa n } n N is convergent and n (λa n = λα. (ii {a n b n } n N is convergent and n (a n b n = α β. (iii If b n 0 for all n N and β 0, then n ( an b n = α β. { an b n }n N is convergent with Proof (i λ, λ, λ,... is a convergent sequence with it λ. Use Theorem 3.7(ii with b n = λ for all n. (ii By part (i with λ =, if {b n } n N is convergent then { b n } n N is convergent. Now use Theorem 3.7(i with {a n } n N and { b n } n N. ( (iii By Theorem 3.7(iii n b n =. Now use Theorem 3.7(ii β { with b n }n N Example Find and {a n } n N. 4n 3 + 3n n 5n. 3 Solution Rough work. For very large n the numerator looks like 4n 3 (imagine evaluating 4n 3 + 3n with n = 0 on your calculator, to the significant figures shown on the calculator you would get the same result evaluating 4n 3 with n = 0 on your calculator. Similarly, for very large n the denominator looks like 5n 3. Hence for very large n the quotient looks like 4n 3 /( 5n 3 = 4/5. So we might guess that the it is 4/5. Next, we might try to use Corollary 3.8(iii to prove this. attempt might be to say 4n 3 + 3n (4n 3 + 3n n 5n = n 3 ( n 5n3. But we would be wrong. This is because Corollary 3.8(iii says that if the two sequences in the quotient converge individually Our first then the it of the quotient is the same as the quotient of the its. In our case the two sequences are {4n 3 + 3n} n N and { 5n 3 } n N, neither 4

5 of which converge. So, before an application of Corollary 3.8 we have to change these two sequences into ones that converge. The usual way is to divide the top and bottom of the quotient by the largest power of n in the bottom term (denominator. In the case, this means dividing top and bottom by n 3 to get sequences {4 + 3/n } n N and {/n 3 5} n N, both of which converge. End of rough work. From an earlier example we know that n = 0. We will use this n fact as follows. 4n 3 + 3n n = n 5n 3 n 5 n( = = = n n n n n ( The idea is to get terms on top and bottom that have finite its, ( 5 n 3 (Corollary 3.8(iii ( 3 n n n ( n n 5 3 n ( (Theorem 3.7 and Corollary n n ( 3 5 (Corollary 3.8(iii = 4 5 = 4 5. See also Question 8 Sheet 3 Example Let {a n } n N be the sequence defined recursively by a = 0, and a n+ = 4 (a n + for all n N. Show that this sequence is convergent and find its it. Solution Rough work. The first few terms are 0,, 5, 4 6 at these we might guess that the sequence is increasing since, ,.... Looking 0 < 4 = 4 6 < 5 6 = 0 64 < 64 = < We might also guess that the sequence is bounded above by, simply because the numerator of each term is smaller than the denominator. Hence we should be thinking of using Theorem

6 End of Rough work (i We will show by induction that a n a n+ holds for all n. If n = then a = 0 and a = in which case a 4 a holds as required. Assume true for n = k, so we are assuming that a k a k+ holds. (We hope to show that a k+ a k+ holds. Then a k a k+, a k + a k+ + + by assumption, adding to each term, (a 4 k + (a 4 k+ + dividing each term by 4, 4 a k+ a k+ < by the inductive definition of the sequence. So, a k+ a k+, i.e. the result is true for n = k +. Thus, by induction, a n a n+ for all n. So, {a n } n N is increasing (since a n a n+ for all n and bounded above by (since a n for all n. Hence, by Theorem 3.4, {a n } n N converges, with it α, say. (ii To find α we note that the sequences 0,, 5,,... and, the same it, i.e. both {a n } n and {a n+ } n N = { (a 4 n + } n N same it. Hence, 6 64,... have have the α = a n = a n+ = n n n 4 (a n + = ( 4 a n + by Theorem 3.7 and Corollary 3.8, n 4 = α From this we see that α = /3 as given in the theorem. Note In this problem we had to choose a number λ and then check that a n λ for all n. Here we chose λ =. How do we find λ? We should do some rough work first, assume that the it exists and run through the argument in part (ii above to find α = /3. We choose any λ /3, usually a simply value such, as in this case,. See also Questions 9 and 0 Sheet 3 6

7 In the proof of theorem 3.0 we will make use of the following result. Lemma 3.9 (Bernoulli s inequality. For all δ and all n we have ( + δ n + nδ. Proof by induction on n. If n = then we get equality. Assume true for n = k, so ( + δ k + kδ. (We now try to show that ( + δ k+ + (k + δ holds. Consider ( + δ k+ = ( + δ( + δ k ( + δ( + kδ by inductive assumption and + δ 0, = + δ + kδ + kδ > + (k + δ, dropping the kδ 0 term. Thus the result holds for n = k +. Hence, by induction, ( + δ n + nδ for all n. See also Question 8 Sheet 4 Theorem 3.0 (i For a fixed integer k, if k < 0, then {n k } n N is convergent and n n k = 0, if k = 0, then {n k } n N is convergent and n n k =, if k > 0, then {n k } n N is divergent. (ii For a fixed real number x, if x <, then {x n } n N is convergent and n x n = 0, if x =, then {x n } n N is convergent and n x n =, if x = or x >, then {x n } n N is divergent. Proof (i (Left to student. (ii Suppose x <. Let ε > 0 be given. If x = 0 then the result is clear. So suppose 0 < x <. In this case we use a TRICK and look at which satisfies >. Write x x x = + δ with δ > 0. By the Archimedean property we can find N N with < δε. Then, N for all n N, 7

8 x n 0 = x n = ( + δ, n + nδ by Lemma 3.9, < nδ Nδ < δε δ = ε. Therefore n x n = 0. Suppose x = then x n = for all n and n x n = is immediate. Suppose x = then the sequence is,,,,,,... which alternates and does not converge, i.e. it diverges. Suppose x >. Write x = + δ with δ > 0. Let λ > 0 be given. By the Alternative Archimedean property, Theorem., we can find N N with N > λ/δ. Then x N = ( + δ N + Nδ by Lemma 3.9, Nδ > λ δ δ = λ. Thus λ is not an upper bound for the sequence {x n } n N. But λ was arbitrary so there is no upper bound for {x n } n N, i.e. the sequence is unbounded. Therefore, by Corollary 3.3, the sequence is divergent. Theorem 3. (Sandwich Rule Let {b n } n N, {c n } n N be convergent sequences with the same it l and b n c n for all n N. Suppose {a n } n N satisfies b n a n c n for all n N. Then {a n } n N is convergent with it l. Proof Let ε > 0 be given. Since{b n } n N is convergent with it l there exists N N such that b n l ε for all n N. This means l ε < b n < l + ε. In particular, l ε < b n ( for all n N. Since{c n } n N is convergent with it l there exists N N such that c n l ε for all n N. In particular, c n < l + ε ( 8

9 for all n N. Let N = max (N, N. Then for all n N both ( and ( hold and so l ε < b n a n c n < l + ε. Thus a n l < ε for all n N as required.. Example Show that cos ( π n 4 = 0. n n Solution For all n we have ( π cos 4 n and so n cos ( π 4 n n. Thus the terms of our sequence are sandwiched by the sequences n and both of which have it 0. Hence, Theorem 3. gives the result. n Note We might try to use Theorem 3.7(ii and say ( cos n ( ( π = cos n n n n n 4 n ( π = 0 cos 4 n = 0. n ( See also Question Sheet 4 But we would be wrong! This is because Theorem 3.7(ii says that if the two sequences in the product converge individually then the it of the product is the product of the its. In our case the two sequences are {/n } n N and { cos ( π n}, the second 4 n N of which does not converge. Thus we cannot apply Theorem 3.7. Example Show that n + ( n n n ( n =. 9

10 Solution For all n we have n n + n + ( n n ( n n + n. So the terms of our sequence are sandwiched by the sequences /n +/n and +/n both of which have it. Hence, Theorem 3. gives the result. /n Appendix Theorem 3.0 (i For a fixed integer k, if k < 0, then {n k } is convergent and n n k = 0, if k = 0, then {n k } is convergent and n n k =, if k > 0, then {n k } is divergent. Proof (Not for examination. (i Suppose k < 0. Let ε > 0 be given. By the Archimedean property we can find N N with N < ε. Since k < 0 and k Z we must, in fact, have k and thus n k n for all n. Hence, for any n N we have n k 0 = n k n N < ε. Therefore n n k = 0. If k = 0 then n k = n 0 = for all n and so n n k =. Suppose k > 0. Let λ > 0 be given. By the Alternative Archimedean property, Theorem., we can find N N with N > λ. Since k > 0 and k Z we must, in fact, have k in which case N k N > λ. Thus λ is not an upper bound for {n k }. But λ was arbitrary so there is no upper bound for {n k }, i.e. the sequence is unbounded. Therefore, by Corollary 3.3, the sequence is divergent. Sufficiently large. We say that a property, p (n, that depends on a natural number n, holds for all sufficiently large n, if there exists N N such that p (n holds for all n N. There are sufficiently large variants of most of our results. For instance I leave it to the student to give a proof of the following: Let {b n } n N, {c n } n N be convergent sequences with the same it l. Suppose {a n } n N satisfies b n a n c n for all sufficiently large n then {a n } n N is convergent with it l. 0

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